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Two players, A and B, alternately and independently flip a biased coin and the first player to get a head wins. Assume player A flips first. Player A wins the game 12/23 times. If the coin is biased, what is the bias of the coin?

I am using the format from here Two players alternately flip a coin; what is the probability of winning by getting a head? except my equation looks 12/23 = p + (1-p)(11/23) and solving for p. I am getting p = 1/12.

I am not understanding the answer or if it is correct. If player A is more likely to win and has first flip, why is the chance of getting heads 1/12??

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1 Answer 1

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Assume the probability of heads is $p$. Now, let the probability that A wins be $P(A)$.

$$ P(A) = p + (1-p)^2p + ... $$ $$ \implies P(A) = \frac{p}{1 - (1-p)^2} = \frac{p}{2p - p^2} = \frac{1}{2-p}$$

Now, equating $P(A)$ to $\frac{12}{23}$, we get

$$ 2-p = \frac{23}{12} \implies p = \frac{1}{12}$$

So, yes, your answer seems correct. This method serves as a verification.

Now, to answer your question keep in mind that if $A$ goes first, it will have higher chance of winning no matter what the bias of the coin is.

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  • $\begingroup$ Can you explain how you went from the summation to 1-(1-p)^2? $\endgroup$
    – user2202
    Commented Oct 28, 2017 at 21:22
  • $\begingroup$ Infinite GP sum $\endgroup$
    – Siddhartha
    Commented Oct 28, 2017 at 21:24
  • $\begingroup$ What is GP? I blieve it is summation but what does GP stand for? $\endgroup$
    – user2202
    Commented Oct 29, 2017 at 3:37
  • $\begingroup$ Geometric progression, sorry I assumed you were familiar $\endgroup$
    – Siddhartha
    Commented Oct 29, 2017 at 5:36

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